Lem. II

Lem. II. The motion or velocity acquired by a ball, in freely descending from rest, by the force of a uniform gravity, is as the time of the descent; and the space fallen through, as the square of that time.

The first part of this lemma is extremely obvious: for since every motion is proportional to the force whereby it is generated, that generated by the force of an uniform gravity must be as the time of the descent; because the whole effort of such a force is proportional to the time of its action; that is, as the time of the descent.

To demonstrate that the distances descended are proportional to the squares of the times, let the time of falling through any proposed distance AB, be represented by the right line PQ; which conceive to be divided into an indefinite number of very small, equal particles, represented each by the symbol $m$; and let the distance descended in the first of them be $Ac$; in the second $cd$; in the third $de$; and so on.

Then the velocity acquired being always as the time from the beginning of the descent, it will at the middle of the first of the said particles be represented by $\backslash frac\{1\}\{2\}m$; at the middle of the second, by $1\backslash frac\{1\}\{2\}m$; at the middle of the third, by $2\backslash frac\{1\}\{2\}m$, &c. which values constitute the series $\backslash frac\{m\}\{2\},\; \backslash frac\{3m\}\{2\},\; \backslash frac\{5m\}\{2\},\; \backslash frac\{7m\}\{2\},\; \backslash frac\{9m\}\{2\}$, &c.

But since the velocity, at the middle of any one of the said particles of time, is an exact mean between the velocities of the two extremes thereof, the corresponding particle of the distance AB may be therefore considered as described with that mean velocity; and so the spaces $Ac,\; cd,\; de,\; ef$, &c. being respectively equal to the abovementioned quantities $\backslash frac\{m\}\{2\},\; \backslash frac\{3m\}\{2\}$,

$\backslash frac\{5m\}\{2\},\; \backslash frac\{7m\}\{2\}$, &c. it follows, by the continual addition of these, that the spaces $Ac,\; Ad,\; Ae,\; Af$, &c. fallen thro' from the beginning, will be expressed by $\backslash frac\{m\}\{2\},\; \backslash frac\{4m\}\{2\},\; \backslash frac\{9m\}\{2\},\; \backslash frac\{16m\}\{2\},\; \backslash frac\{25m\}\{2\}$, &c. which are evidently to one another in proportion, as 1, 4, 9, 16, 25, &c. that is, as the squares of the times. Q. E. D.

Corol. Seeing the velocity acquired in any number ($n$) of the aforesaid equal particles of time (measured in the space that would be described in one single particle) is represented by ($n$) times $m$, or $nm$; it will therefore be as one particle of time is to $n$ such particles, so is $nm$, the said distance answering to the former time, to the distance, $n^2m$, corresponding to the latter, with the same celerity acquired at the end of the said $n$ particles. Whence it appears, that the space $\backslash frac\{n^2m\}\{2\}$ (found above) through which the ball falls in any given time $n$, is just the half of that time ($n^2m$) which might be uniformly described with the last, or greatest celerity in the same time.

Schol. It is found by experiment, that any heavy body near the earth's surface (where the force of gravity may be considered as uniform) descends about 16 feet from rest, in the first second of time. Therefore, as the distances fallen through are proved above to be in proportion as the squares of the time, it follows, that as the square of one second is to the square of any given number of seconds, so is 16 feet to the number of feet a heavy body will freely descend in the said number of seconds. Whence the number of feet descended in any given time will be found, by multiplying the square of the number of seconds by 16. Thus the distance descended in 2, 3, 4, 5, &c. seconds, will appear to be 64, 144, 256, 400 feet, &c. respectively. Moreover, from hence, the time of the descent through any given distance will be obtained, by dividing the said distance in feet by 16, and extracting the square root of the quotient; or, which comes to the same thing, by extracting the square root of the whole distance, and then taking $\backslash frac\{1\}\{2\}$ of that root for the number of seconds required. Thus, if the distance be supposed 2640 feet; then, by either of the two ways, the time of the descent will come out 12.84, or 12.50 seconds.

It appears also (from the corol.) that the velocity per second (in feet) at the end of the fall, will be determined by multiplying the number of seconds in the fall by 32. Thus it is found, that a ball, at the end of 10 seconds, has acquired a velocity of 320 feet per second. After the same manner, by having any two of the four following quantities, viz. the force, the times, the velocity, and distance, the other two may be determined; for let the space freely descended by a ball, in the first second of time (which is as the accelerating force) be denoted by $F$; also let $T$ denote the number of seconds wherein any distance, $D$, is descended; and let $V$ be the velocity per second, at the end of the descent: then will $V=2FT=2\backslash sqrt\{FD\}=2DT=\backslash sqrt\{D\}=V=$

Projectiles. All which equations are very easily deduced from the two original ones, $D=FTT$, and $V=2FT$, already demonstrated; the former in the proposition itself, and the latter in the corollary to it; by which it appears, that the measure of the velocity at the end of the first second is $2F$; whence the velocity ($V$) at the end of ($T$) seconds must consequently be expressed by $2FT$ or $2TT$.

Theorem 1. A projected body, whose line of direction is parallel to the plane of the horizon, describes by its fall a parabola. If the heavy body is thrown by any extrinsic force, as that of a gun or the like, from the point $A$, (fig. 2. n^o 1.) so that the direction of its projection is the horizontal line $AD$; the path of this heavy body will be a semi-parabola. For if the air did not resist it, nor was it acted on by its gravity, the projectile would proceed with an equable motion, always in the same direction; and the times wherein the parts of space $AB$, $AC$, $AD$, $AE$, were passed over, would be as the spaces $AB$, $AC$, $AD$, &c. respectively. Now if the force of gravity is supposed to take place, and to act in the same tenour as if the heavy body were not impelled by any extrinsic force, that body would constantly decline from the right-line $AE$; and the spaces of descent, or the deviations from the horizontal line $AE$, will be the same as if it had fallen perpendicularly. Wherefore if the body falling perpendicularly by the force of its gravity, passed over the space $AK$ in the time $AB$, descended through $AL$ in the time $AC$, and through $AM$ in the time $AD$; the spaces $AK$, $AL$, $AM$, will be as the squares of the times, that is, as the squares of the right-lines $AB$, $AC$, $AD$, &c. or $KF$, $LG$, $MH$. But since the impetus in the direction parallel to the horizon always remains the same, (for the force of gravity, that only solicits the body downwards, is not in the least contrary to it), the body will be equally promoted forwards in the direction parallel to the plane of the horizon, as if there was no gravity at all. Wherefore, since in the time $AB$, the body passes over a space equal to $AB$; but being compelled by the force of gravity, it declines from the right-line $AE$ through a space equal to $AK$; and $BF$ being equal and parallel to $AK$, at the end of the time $AB$, the body will be in $F$; so in the same manner, at the end of the time $AE$, the body will be in $I$, and the path of the projectile will be in the curve $AFGHI$; but because the squares of the right lines $KF$, $LG$, $MH$, $NI$, are proportionable to the abscisses $AK$, $AL$, $AM$, $AN$. The curve $AFGHI$ will be a semi-parabola. The path, therefore, of a heavy body projected according to the direction $AE$, will be a semi-parabolic curve.

Theorem 2. The curve line that is described by a heavy body projected obliquely and upwards, according to any direction, is a parabola.

Let $AF$ (ibid. n^o 2.) be the direction of projection, any ways inclined to the horizon, gravity being supposed not to act, the moving body would always continue its motion in the same right-line, and would describe the spaces $AB$, $AC$, $AD$, &c. proportional to the times. But by the action of gravity it is compelled continually to decline from the path $AF$, and to move in a curve, which will be a parabola. Let us suppose the heavy body falling perpendicularly in the time $AB$, through the space $AQ$, and in the time $AC$,

through the space $AR$, &c. The spaces $AQ$, $AR$, Projectiles. $AS$, will be as the squares of the times, or as the Projection. squares of $AB$, $AC$, $AD$. It is manifest from what was demonstrated in the last theorem, that if in the perpendicular $BG$, there is taken $BM=AQ$, and the parallelogram be completed, the place of the heavy body at the end of the time $AB$, will be $M$, and so of the rest; and all the deviations $BM$, &c. from the right-line $AF$, arising from the times, will be equal to the spaces $AQ$, $AR$, $AS$, which are as the squares of the right-lines $AB$, $AC$, $AD$. Through $A$ draw the horizontal right-line $AP$, meeting the path of the projectile $P$. From $P$ raise the perpendicular $PE$, meeting the line of direction in $E$; and by reason the triangles $ABG$, $ACH$, &c. are equiangular, the squares of the right-lines $AB$, $AC$, &c. will be proportionable to the squares of $AG$, $AH$, &c. so that the deviations $BM$, $CN$, &c. will be proportionable to the squares of the right lines $AG$, $AH$, &c. Let the line $L$ be a third proportional to $EP$ and $AP$; and it will be (by 17 El. 6.) $L\; \backslash times\; EP\; =\; APq$. but $APq\; :\; AGq\; ::\; EP\; :\; BM\; ::\; L\; \backslash times\; EP\; :\; L\; \backslash times\; BM$; whence since it is $L\; \backslash times\; EP\; =\; APq$. it will be $L\; \backslash times\; BM\; =\; AGq$. In like manner it will be $L\; \backslash times\; CN\; =\; AHq$. &c. But because it is $BG\; :\; AG\; ::\; (EP\; :\; AP\; ::\; \backslash text\{by\; hypothesis\})\; AP\; :\; L$; it will be $L\; \backslash times\; BG\; =\; AG\; \backslash times\; AP\; =\; AG\; \backslash times\; AG\; +\; AG\; \backslash times\; GP\; =\; AGq\; +\; AG\; \backslash times\; GP$. But it has been shown that it is $L\; \backslash times\; BM\; =\; AGq$. wherefore it will be $L\; \backslash times\; BG\; -\; L\; \backslash times\; BM\; =\; AG\; \backslash times\; GP$, that is, $L\; \backslash times\; MG\; =\; AG\; \backslash times\; GP$. By the same way of reasoning it will be $L\; \backslash times\; NH\; =\; AH\; \backslash times\; HP$, &c. Wherefore the rectangle under $MG$ and $L$, will be equal to the square of $AG$, which is the property of the parabola; and so the curve $AMNOPK$ wherein the projectile is moved will be a parabola.

Cor. 1. Hence the right line $L$ is the latus rectum or parameter of the parabola, that belongs to its axis.

Cor. 2. Let $AH=HP$, and it will be $L\; \backslash times\; CN\; =\; AHq\; =\; L\; \backslash times\; NH$, whence it will be $NH=CN$; and consequently the right-line $AF$ being the line of direction of the projectile, will be a tangent to the parabola.

Cor. 3. If a heavy body is projected downwards, in a direction oblique to the horizon, the path of the projectile will be a parabola.

Theorem 3. The impetus of a projected body in different parts of the parabola, are as the portions of the tangents intercepted betwixt two right-lines parallel to the axis; that is, the impetus of the body projected in the points $A$ and $B$, (ibid. n^o 3.) to which $AD$ and $BE$ are tangents, will be as $CD$ and $EB$, the portions of the tangents intercepted betwixt two right-lines, $CB$ and $DE$, parallel to the axis.

These calculations and demonstrations, however, are all founded on a supposition that the projectiles move in an unresisting medium, or in one whose resistance is but small. Hence they answer with tolerable exactness where the motions are not very quick; but in those cases where the projectiles are moved with immense velocity, the resistance of the air occasions errors of such enormous magnitude, that a musket-ball, which, by calculation, ought to fly 17 miles, seldom exceeds three quarters of a mile. See the article GUNNERY, passim.